Electrochemistry 1 Question 18
18. The standard reduction potentials $E^{\circ}$, for the half reactions are as
$ \begin{aligned} \mathrm{Zn} & =\mathrm{Zn}^{2+}+2 e^{-}, E^{\circ}=+0.76 \mathrm{V} \\ \mathrm{Fe} & =\mathrm{Fe}^{2+}+2 e^{-}, E^{\circ}=0.41 \mathrm{V} \end{aligned} $
The emf for the cell reaction,
$ \mathrm{Fe}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \text { is } $
$(1989,1 \mathrm{M})$
(a) $-0.35 \mathrm{V}$
(b) $+0.35 \mathrm{V}$
(c) $+1.17 \mathrm{V}$
(d) $-1.17 \mathrm{V}$
Show Answer
Answer:
Correct Answer: 18. (b)
Solution:
- $\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; \quad E^{\circ}=-0.41 \mathrm{V}$
$ \mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-} ; E^{\circ}=+0.76 \mathrm{V} $
$\Rightarrow \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} ; \quad E^{\circ}=+0.35 \mathrm{V}$
